Example 3.2 Flat slabs - · PDF fileExample 3.2 Flat slabs The following floor plan is...

Addis Ababa Institute of Technology School of Civil and Environmental Engineering Reinforced Concrete Structures II

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Example 3.2 Flat slabs

The following floor plan is intended to be a flat slab system. The slab thickness is 200mm thick and

supports a characteristic dead load of 2.7 KN/m2 in addition to self-weight and a characteristic live load

of 3 KN/m2. The slab is provided on edge beams of bw= 300 mm and h=400mm and all columns are

300mm by 300mm. Design the slab system using C20/25, S – 400.

Solution:

Step 1: Material property

Concrete:

𝑓𝑐𝑑 = 0.85 ∗ 20

1.5= 11.33𝑀𝑝𝑎

𝑓𝑐𝑡𝑘, 0.05 = 1.5𝑀𝑝𝑎

𝑓𝑐𝑡𝑚 = 2.2𝑀𝑝𝑎

ɤ𝑐 = 1.5


2

𝑓𝑐𝑘 = 20𝑀𝑝𝑎, 𝑓𝑐𝑢 = 25𝑀𝑝𝑎

Rebar: 𝑓𝑦𝑘 = 400𝑀𝑝𝑎

𝑓𝑦𝑑 =𝑓𝑦𝑘

1.15= 347.83𝑀𝑝𝑎

ɛ𝑦𝑑 =𝑓𝑦𝑑

𝐸𝑠=

347.83

200= 1.74%0

Step 2: Depth check for deflection

According to ACI code minimum thickness of slabs without interior beams: without drop

panels for fyk = 400 Mpa

- Exterior panels with edge beams = ln /33

Slab considered to have edge beam if 0.8bf

s

I

I

for this case 8

8

20.33 101.05 0.8 !

19.33 10

bf

s

I xok

I x

(see the computation of Ib and Is on page 6)

- Interior panels = ln /33

Ln is the length of the clear span in longer direction.

- For edge panel Ln = 5500 - 300 = 5200mm

- For edge panel Ln = 6000 - 300 = 5700mm

5700172.73

33 33

nld mm

12cov 172.73 15 193.7

2 2h d er mm

<200 mm ok!

Step 3: Loading

Dead load:

Self-weight → 0.2 * 25 = 5 KN/m2

Imposed dead load → 2.7 KN/m2

Gk = 7.7 KN/m2

Variable Load:

Live load Qk = 3 KN/m2

Design load:

𝐺𝑑 = 1.35 ∗ 𝐺𝑘 = 1.35 ∗ 7.7 = 10.39 KN/m2

𝑄𝑑 = 1.5 ∗ 𝑄𝑘 = 1.5 ∗ 3 = 4.5 KN/m2

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑃𝑑 =14.89 KN/m2


3

Step 4: Limitations to direct design method

1. 3 span in each direction…………. Ok!

2. 5.5 6

1.15 2 1.25 2........ !4.8 4.8

y

x

Land Ok

L

3. Successive span length in each direction shall not differ by more than 1/3 of longer span.

16 5.5 0.5 6 2 !

3x Ok

4. All loads must be due to gravity only …… (By assuming there are no lateral loads on our

slab) Ok!

5. Factored live load must be less than twice of the factored dead load

4.52 0.43 2 !

10.395

LLOk

DL

6. Maximum offset of column from either axis between center-line of successive columns

shall not exceed 10% of the span in direction of offset …. Ok!

7. For a panel with beams between supports on all sides the relative stiffness of the beams

in the two perpendicular direction given by 2 2

1 2 2 1( . ) / ( . ) Shall not be less than 0.2 or greater than 5 …. Ok!

The direct design method (DDM) can be used.

Step 5: Analysis


4

Due to symmetry strip-1 & strip-4 are similar and

Strip -2 & strip 3 are similar.

Strip -1 (along axis -1)

i. Compute l2, 2

4.8 0.32.4 0.15 2.55

2 2mm

ii. Compute Ln,𝑙𝑛 = {5500 − 300 = 5200𝑚𝑚

𝑏

𝑛 𝑎𝑥𝑒𝑠 𝐴 𝑎𝑛𝑑 𝐵 , 𝐶 𝑎𝑛𝑑 𝐷

6000 − 300 = 5700𝑚𝑚𝑏

𝑛 𝑎𝑥𝑒𝑠 𝐵 𝑎𝑛𝑑 𝐶

iii. Compute lx,𝑙𝑥 = 4.8𝑚 The minimum panel dimensions

𝑙𝑥 =4.8

4= 1.2

iv. Compute static moments

2

2

8

d no

P l lM

Between axis A & B or C & D 2

0

14.89 2.55 5.2128.34 .

8

x xM KN m

Between axis B & C 2

0

14.89 2.55 5.7154.2 .

8

x xM KN m

v. Longitudinal distribution of Mo

a. Between axis A & B or C & D

128.34 .oM KN m

N. B:-It is case-4, slabs without beam between interior supports and with edge beam.


5

- For panel between axis A and B

- For panel between axis C and D

b. Between axis B &C

153.38 .oM KN m

N.B Since there is a beam on the column strip some amount of the column strip moment will be

resisted by the beam

4

4

bcb

fb

cb

IE

lI

El

Since “l” is the same for both the beam and slab and cb csE E

bf

s

I

I


6

400

200

200

300

min( ,4 ) 200

h mm

hs mm

hw mm

bw mm

t hw hs mm

(500 200) 300 (300 200) 100225

(500 200) (300 200)

x x x xy mm

x x

3

3

9 6

1

12

12550 200

12

1.7 10 1700 10

sI bh

x x

x x mm

3 2 3 21 1300 200 (300 200)(225 100) 500 200 (500 200)(175 100)

12 12bI x x x x x x

6 6

6 4

1137.5 10 895.83 10

2033.33 10

x x

x mm

3 6 412550 200 1700 10

12sI x x x mm

6

6

2033.33 101.196

1700 10

b

s

I x

I x

Calculate 2

1

( )f

l

l

- 2 & &4 .8 For panel between axis A B or C Dm

- 1 6.0m for panel between axis B & C.


7

- For panel between axis A & B and C & D

2

1

4.8( ) 1.196( )

5.5f

1.045 1.0

- For panel between axis B & D

2

1

4.8( ) 1.196( )

6f

0.957 1.0

vi. Transverse distribution of moments (i.e. to column and middle strip)

For panel between axis A & B and C & D

a. Interior negative moment beam, M=89.84 KN.m

- There is intermediate beam

2

1

( ) 1.0f

2

1

4.80.87

5.5

- There is intermediate beam, 78.9 % (interpolated form table)

C.S moment 0.789 89.84 70.88 .x KN m

(1 0.789) 0.211

Half M.S moments 0.211 89.84 18.96 .x KN m

b. Positive moment, M=64.17 KN.m

2

1

( ) 1.0f

2

1

4.80.87

5.5

- 78.9% of moment to column strip

C.S moment 0.789 64.17 50.63 .x KN m

Half M.S moments 0.211 64.17 13.54 .x KN m

c. Exterior negative moment, M=38.50 KN.m

- 2

1

( ) 1.0f


8

- There is edge beam 0t

Calculate 2 s

c

I

3

[(1 0.63 )* ]3

.

x x yc

y

N B Always x y

Case 1:

3 3

1

3 4

0.2 0.2 *0.5 0.2 0.2 *0.3[(1 0.63* )*( )] [(1 0.63* )*( )]

0.5 3 0.3 3

(0.748*0.00133) (0.58 0.0008)

0.00097 0.00046

1.43*10

c

m

Case 2:

3 3

1

3 4

0.2 0.2 *0.3 0.2 0.3 *0.4[(1 0.63* )*( )] [(1 0.63* )*( )]

0.2 3 0.4 3

(0.37*0.00053) (0.5275 0.0036)

0.0002 0.00140

2.10*10

c

m

1 2

3 4

max ,

2.10*10

c c c

c m

- Calculate Is


9

3

3

3 4

1* *

12

1*2.55*0.2

12

1.7*10

s

s

I b h

I

m

3

3

2.1*10

2* 2*1.7*10

0.618

t

s

C

I

Percentage of column strip moment can be determined by interpolation.

1

1

220.618, 1, 0.5 97.5281 1

220.618, 1, 1 93.821 1

20.872 94.76

1

t f

t f

llfor andl l

llfor andl l

lfor and thevalues obtained above

l

. 0.948*38.5

36.5

. 0.052*38.5

2.00

C S moment

KNm

M S moment

KNm

For panel between axis B & C

a) Interior negative moment, M = 100.23 KN.m

1

1

1

22 0, 0.8 751 1

22 1, 0.8 811 1

2 0.957, 80.7421

f

f

f

llinerpoliate for andl l


linerpoliate for and thevalues obtained abovel

. 0.8074*100.23

80.93

. 0.1926*100.23

19.3

C S moment

KNm

M S moment

KNm

b) Positive moment, M = 53.97 KN.m


10

1

1

1

22 0, 0.8 601 1

22 1, 0.8 811 1

2 0.957, 80.0971

f

f

f



linerpoliate for and thevalues obtained abovel

. 0.801*53.97

43.23

. 0.194*53.97

10.74

C S moment

KNm

M S moment

KNm

Note: If 2

1

( ) 1.0f

l

l 85% of the column strips moment goes to the beam and 15% to the slab.

- For panel between axis A & B or C & D, 85% of the column strip moment goes to

beam and 15% to the slab.

- For panel between axis B & C

>= 1.0 85%

0.957 x

0.0 0%

0.957 0.0 1.0 0.0

0 85 0x

0.957 8581.35%

1

xx

Therefore, 81.35% of column strip moment goes to the beam and 18.65% goes to the

slab.

- For panel between axis A & B and C & D

Moment Location Value To beam

(85%)

To slab

(15%)

Exterior –ve moment 36.5 31.0 5.5

Positive moment 50.63 43.04 7.59

Interior –ve moment 70.88 60.248 10.632

- For panel between axis B & C

Moment Location Value To beam To slab

Interior –ve moment 80.93 65.43 15

Positive moment 43.23 35.17 8.06


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Moment summary

Strip 2: [along axis 2]

i) Compute l2,

4.8 4.822 2

4.8

l

ii) Compute ln,

ln 5500 300 5200 / & &

6000 300 5700 / &

mmb naxis A B and C D

mmb naxis B C

iii) Compute lx, min 2, n 4.8lx l l m


12

1.22

lxhalf a weidthof thecolumn strip m

iv) Compute the static moment, Mo. 2

2* * n

8

do

P l lM

Between axis A & B or C & D 214.89*4.8*5.2

8

241.58

oM

KN m

Between axis A & B or C & D 214.89*4.8*5.7

8

290.26

oM

KN m

v) Longitudinal distribution of Mo = 241.58KN-m

a) Between axis A & B or C & D

- For panel between axis A and B,

- For panel between axis C and D


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b. Between axis B &C M0 =290.26KN-m

vi) Transverse distribution of moments (i.e to column and middle strip)

For panel between axis A & B and C & D

a) Interior negative moment, M=169.1 KN-m

No intermediate beam → α=0

75% of moment to column strip

. 0.75*169.1

126.8

. 0.125*169.1

21.14

C S moment

KNm

half M S moment

KNm

b) Positive moment, M=120.79 KN-m



. 0.6*120.79

72.47

. 0.2*120.79

24.16

C S moment

KNm

half M S moment

KNm

c) Exterior negative moment, M=72.47 KN-m


There is edge beam → βt≠0

calculate2*

t

s

C

I

3 42.10*10 ( )c m form previouscalculattake ion


14

3

3

3 4

1* *

12

1*4.8*0.2

12

3.2*10

s

s

I b h

I

m

3

3

2.1*10

2* 2*3.2*10

0.328

t

s

C

I

Percentage of column strip moment can be determined by interpolation

0 100%

0.328

2.5 75%

96.72%

t

t

x

x

. 0.9672*72.47

70.09

. 0.0164*72.47

1.188

C S moment

KNm

half M S moment

KNm

For panel between axis B & C

a) Interior negative moment, M = 188.67KN-m



. 0.75*188.67

141.5

. 0.125*188.67

23.58

C S moment

KNm

half M S moment

KNm

b) positive moment, M = 101.6KN-m




15

. 0.6*101.6

60.9

. 0.2*101.6

20.32

C S moment

KNm

half M S moment

KNm

Moment summary

Step 6: Design

cov2

10200 152

180

d h er for thebottomlayer bar

mm

Area of minimum reinforcement

,min

2

,min

0.26max ,0.0013

0.26*2.2*1000*180max ,0.0013*1000*180

400

max 257.4,234

257.4

ctm ts

yk

s

f b dA bd

f

A mm


16

Minimum spacing:

min max , 5 ,20S aggrigate size mm mm

Maximum spacing:

max

max

min 3 ,400 , f

0

in

40

S h mm for primary re orcement

S mm

Moment diagram (not drawn to scale)


17

strip Moment strip width

moment per meter ω As,req s, req s provided

C.S 5.5 1.35 4.07 0.011 257.4 333 use ф 10 C/C300

7.59 1.35 5.62 0.015 257.4 333 use ф 10 C/C300

10.632 1.35 7.88 0.0216 257.4 333 use ф 10 C/C300

15.1 1.35 20.39 0.057 332.32 250 use ф 10 C/C300

8.06 1.35 10.88 0.03 257.4 333 use ф 10 C/C300

M.S 18.96 1.2 15.80 0.044 257.4 333 use ф 10 C/C300

13.54 1.2 11.28 0.317 257.4 333 use ф 10 C/C300

2 1.2 1.67 0.0045 257.4 333 use ф 10 C/C300

19.3 1.2 16.08 0.0446 261.9 333 use ф 10 C/C300

10.74 1.2 8.95 0.024 257.4 333 use ф 10 C/C300

C.S 126.8 2.4 52.63 0.157 921.99 90.9 use ф 10 C/C90

72.47 2.4 30.03 0.0889 521.19 166.67 use ф 10 C/C160

70.09 2.4 29.20 0.083 45.5 166.67 use ф 10 C/C160

141.5 2.4 58.96 0.177 1036.99 76.9 use ф 10 C/C70

60.9 2.4 25.40 0.072 422.8 200 use ф 10 C/C200

M.S 21.14 1.2 17.54 0.0487 286.01 333 use ф 10 C/C300

24.16 1.2 20.03 0.0559 327.9 250 use ф 10 C/C250

1.188 1.2 0.99 0.0027 257.4 333 use ф 10 C/C300

23.58 1.2 19.65 0.0559 327.9 250 use ф 10 C/C250

20.32 1.2 16.93 0.047 277.39 333 use ф 10 C/C300


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